Wait, no, the user is asking me to solve the problem step by step. Alright, let me try to figure this out.
The problem says that there's a polynomial with integer coefficients, distinct integer roots. The leading coefficient $ a_n $ is 2, and the constant term $ a_0 $ is 66. We need to find the least possible value of $ |a_{n-1}| $.
First, I remember that for a polynomial with integer roots, if the roots are integers, then the polynomial can be written as $ f(x) = 2(x - r_1)(x - r_2)...(x - r_n) $, where each $ r_i $ is an integer. Since the leading coefficient is 2, that means the polynomial is monic multiplied by 2, but actually, the leading term is 2x^n, so when you expand the product, the coefficient of x^n is 2. Therefore, the roots are integers, and the polynomial is factored as 2 times the product of (x - r_i).
But the constant term $ a_0 $ is 66. The constant term of the polynomial is the product of the roots multiplied by the leading coefficient. Wait, let me recall: for a monic polynomial, the constant term is (-1)^n times the product of the roots. But here, the leading coefficient is 2, so the constant term is 2 times (-1)^n times the product of the roots. So:
$ a_0 = 2 \times (-1)^n \times r_1 r_2 \dots r_n = 66 $
Therefore, $ (-1)^n \times r_1 r_2 \dots r_n = 33 $. Since 33 is positive, then either n is even (so that (-1)^n is 1) and the product of roots is 33, or n is odd and the product of roots is -33.
But since the roots are distinct integers, we need to find integers r_1, r_2, ..., r_n such that their product is either 33 or -33, depending on the parity of n, and then compute the coefficient $ a_{n-1} $, which is the coefficient of x^{n-1} in the polynomial.
The coefficient $ a_{n-1} $ is related to the sum of the roots multiplied by (-1)^{n-1} times the leading coefficient. Wait, more precisely, for a polynomial $ f(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_0 $, the coefficient $ a_{n-1} $ is equal to $ (-1)^{1} a_n \times S_1 $, where $ S_1 $ is the sum of the roots. Wait, no, more accurately, using Vieta's formula, the coefficient $ a_{n-1} $ is equal to $ (-1)^{1} \times a_n \times (r_1 + r_2 + \dots + r_n) $. Wait, let me check:
For a monic polynomial $ (x - r_1)(x - r_2)...(x - r_n) = x^n - (r_1 + r_2 + ... + r_n)x^{n-1} + ... + (-1)^n r_1 r_2 ... r_n $. So, if the polynomial is $ 2(x - r_1)...(x - r_n) $, then the coefficient of x^{n-1} is $ -2(r_1 + r_2 + ... + r_n) $. Therefore, $ a_{n-1} = -2 \times S_1 $, where $ S_1 $ is the sum of the roots. Therefore, the absolute value $ |a_{n-1}| = 2 |S_1| $.
So, our goal is to minimize $ |S_1| $, given that the product of the roots is either 33 or -33, and the roots are distinct integers.
So, first, let's consider that the product of the roots is either 33 or -33. Let me think about the factors of 33 and -33. The prime factors of 33 are 3 and 11, so the positive divisors are 1, 3, 11, 33. The negative divisors would be -1, -3, -11, -33. So, the possible integer roots (distinct) that multiply to 33 or -33.
But we need to find all possible sets of distinct integers whose product is 33 or -33, and then compute the sum of the roots, then take absolute value multiplied by 2, and find the minimal such value.
But first, we need to figure out how many roots there are. Since the polynomial is of degree n, and the leading coefficient is 2, but the number of roots is n. However, the problem doesn't specify the degree. So, the degree could be any number, as long as the roots are distinct integers, their product is 33 or -33, and the leading coefficient is 2.
Therefore, we need to find all possible sets of distinct integers (with product 33 or -33) and for each possible set, compute the sum of the roots, then compute 2 times that absolute value, and find the minimal value.
But since the problem asks for the least possible value of |a_{n-1}|, which is 2 times the absolute value of the sum of the roots, we need to minimize |sum of roots|.
Therefore, our task reduces to finding a set of distinct integers with product ±33, such that the sum of the integers is as small as possible in absolute value.
So, the question is: what is the minimal |S| where S is the sum of distinct integers whose product is ±33.
So, let's consider possible factorizations of 33 and -33 into distinct integers.
First, let's think about positive product 33. The possible factorizations into distinct integers:
Since 33 is 1×3×11. That's three distinct integers. Alternatively, 33 can be written as 33 itself (but that's only one number). But since we need distinct integers, if we have more than one root, we need to split 33 into multiple factors. For example, 1×3×11, or maybe 1×(-1)×(-3)×11, but that would be product 33. Wait, but the product could be positive or negative.
Alternatively, maybe we can have more factors by including 1 and -1, but then the product would change.
But let's think step by step.
First, for product 33. The possible distinct integers whose product is 33:
Case 1: Three factors: 1, 3, 11. Product is 33. Sum is 1 + 3 + 11 = 15.
Case 2: Maybe with more factors. For example, including 1, -1, and then other numbers. Let's see. If we have 1, -1, and then 33. The product is 1 × (-1) × 33 = -33. But that's for the product -33. So if we want product 33, maybe we need to have even number of negative factors. For example, two negative numbers and the rest positive. But since 33 is positive, the number of negative factors must be even.
But if we have more than three factors, for example, 1, -1, -3, 11. Then the product is 1 × (-1) × (-3) × 11 = 33. The sum is 1 -1 -3 + 11 = 8. That's better than 15. So that's a smaller absolute sum.
Alternatively, maybe even more factors. For example, 1, -1, 3, -11. Product is 1 × (-1) × 3 × (-11) = 33. Sum is 1 -1 + 3 -11 = -8. Absolute value is 8. Same as before.
Alternatively, maybe 1, -1, 3, -3, 11. Wait, but the product would be 1 × (-1) × 3 × (-3) × 11 = 99. That's too big. Not 33.
Alternatively, 1, -1, 3, -11, and 1. But that would have duplicates. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and something else. Wait, but the product would be 1 × (-1) × 3 × (-11) × x = 33x. So to get 33, x must be 1. But then duplicates. Not allowed.
Alternatively, maybe 1, -1, 3, -3, 11. Wait, product is 1×(-1)×3×(-3)×11 = 99. Not 33. So that's not helpful.
Alternatively, maybe 1, -1, 3, -11, and 1. But again duplicates. Not allowed.
Alternatively, maybe 1, -1, -3, 11. Which is 4 numbers. Product is 33. Sum is 1 -1 -3 + 11 = 8. Alternatively, if we take 1, -1, -3, -11. Then the product is 1 × (-1) × (-3) × (-11) = -33. So that's for the product -33. So that's not helpful for our case.
So, for product 33, with 4 factors, we can have 1, -1, -3, 11. Sum is 8. Alternatively, 1, -1, 3, -11. Sum is -8. So absolute value is 8. So that's better than the three factors case.
Is there a way to get a smaller sum?
What if we have more factors? Let's see. For example, 1, -1, 3, -3, 11. Wait, product is 1×(-1)×3×(-3)×11 = 99. Not 33. Not helpful.
Alternatively, 1, -1, 3, -11, and then something else. Wait, but that would require more factors. Let me think. Let me think of other possibilities.
Alternatively, maybe using more factors by breaking down 33 into more numbers. For example, 33 can be written as 1, -1, 3, -3, 11, but that's product 99. Not helpful. Alternatively, maybe 1, -1, 3, -11, and then 1 again. But duplicates. Not allowed.
Alternatively, maybe 1, -1, 3, -3, 11, -1. Again duplicates.
Alternatively, maybe 1, -1, 3, -11, and then another factor? Let me think. Suppose we have 1, -1, 3, -11, and then another factor. Let's say x. Then the product is 1 × (-1) × 3 × (-11) × x = 33x. To get 33, x must be 1. But that would duplicate 1. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then -1. Again duplicates. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then something else. Maybe 3? But duplicates again.
Alternatively, maybe 1, -1, 3, -3, 11. But that product is 99. Not helpful.
Alternatively, maybe using 1, -1, 3, -11, and then something else. Wait, but that seems not working.
Alternatively, maybe using more negative numbers. For example, 1, -1, -3, -11. Product is 1 × (-1) × (-3) × (-11) = -33. Which is the negative case. So that's not helpful for the positive product.
Alternatively, maybe 1, -1, -3, 11, and -1. Again duplicates.
Alternatively, maybe 1, -1, -3, 11, and 1. Again duplicates.
Alternatively, maybe 1, -1, 3, -11, and then -1. Duplicates again.
So, seems like for the product 33, the minimal |sum| is 8. But maybe there's another way.
Wait, what if we have more factors? For example, 1, -1, 3, -3, 11. Wait, product is 99. Not 33. If I take 1, -1, 3, -3, 11, and divide by something? Not sure.
Alternatively, maybe 1, -1, 3, -11, and then 1. But duplicates.
Alternatively, maybe using more factors but with 1 and -1. For example, 1, -1, 3, -11, and then another 1. No, duplicates.
Alternatively, maybe using 1, -1, 3, -11, and then something else. Let me think. If I have 1, -1, 3, -11, and then x. Then the product is 1×(-1)×3×(-11)×x = 33x. So if x is 1, product is 33. But duplicate 1. If x is -1, product is -33. So that's not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. But I don't think that helps. So maybe 8 is the minimal for product 33.
Now, let's consider the case where the product is -33. Then, the product of the roots is -33. So, similar approach.
So, the roots are distinct integers whose product is -33. Let's think of possible factorizations.
For example, 1, -1, 3, 11. Product is 1×(-1)×3×11 = -33. Sum is 1 -1 + 3 + 11 = 14. Absolute value is 14.
Alternatively, 1, -1, -3, 11. Product is 1×(-1)×(-3)×11 = 33. Not -33.
Alternatively, 1, -1, 3, -11. Product is 1×(-1)×3×(-11) = 33. Not -33.
Alternatively, 1, -1, 3, -11, and then something else? Wait, if we have 1, -1, 3, -11, and then another factor. Let me see. Suppose we have 1, -1, 3, -11, and then x. The product is 1×(-1)×3×(-11)×x = 33x. To get -33, x must be -1. But then duplicate -1. Not allowed.
Alternatively, 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and then 1. Duplicate.
Alternatively, maybe 1, -1, 3, 11, and then -1. Duplicate.
Alternatively, maybe 1, -1, 3, 11, and -1. Duplicate.
Alternatively, maybe 1, -1, -3, -11. Product is 1×(-1)×(-3)×(-11) = -33. Sum is 1 -1 -3 -11 = -14. Absolute value is 14.
Alternatively, maybe more factors. For example, 1, -1, -3, 11, and then something else. Let's see: 1×(-1)×(-3)×11×x = 33x. To get -33, x = -1. But duplicate -1. Not allowed.
Alternatively, 1, -1, -3, 11, and then -1. Duplicate again.
Alternatively, 1, -1, -3, -11, and 1. Duplicate 1.
Alternatively, 1, -1, -3, -11, and then something else. Let me think. If we have 1, -1, -3, -11, and then x. The product is 1×(-1)×(-3)×(-11)×x = -33x. To get -33, x must be 1. So duplicate 1. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and -1. Duplicate -1.
Alternatively, maybe 1, -1, -3, 11, and -1. Duplicate -1.
Alternatively, maybe using more factors. For example, 1, -1, 3, -11, and then -1. Duplicate. Not allowed.
Alternatively, maybe 1, -1, -3, 11, and then -1. Again duplicate.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, -3, 11, and then another factor. Let me try to think of another way.
Alternatively, maybe 1, -1, 3, -11, and then something. If we have 1, -1, 3, -11, and then 1. Duplicate. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then -1. Duplicate.
Alternatively, maybe 1, -1, 3, -11, and then - something else. Wait, if we have 1, -1, 3, -11, and then -x. Then the product is 1×(-1)×3×(-11)×(-x) = -33x. To get -33, we need -33x = -33, so x = 1. Then the roots would be 1, -1, 3, -11, -1. But duplicate -1. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then something else. Let me think. Suppose we have 1, -1, 3, -11, and then something else. Let me think of another factor. For example, 1, -1, 3, -11, and then 1. Duplicate. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then - something. Let me think. If I have 1, -1, 3, -11, and -1. Duplicate again.
Alternatively, maybe 1, -1, 3, -11, and then something else. Maybe something like 1, -1, 3, -11, and then -1. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then - something. Wait, but that's not working.
Alternatively, maybe 1, -1, 3, -11, and then something else. Maybe 1, -1, 3, -11, and then 3. Duplicate.
Alternatively, maybe 1, -1, 3, -11, and then -3. Then the product is 1×(-1)×3×(-11)×(-3) = -99. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
So, for the case of product -33, the minimal sum is 14 or -14. So absolute value is 14. Which is worse than the previous case where we had 8.
So, so far, for product 33, we have a sum of 8 or -8, which gives |a_{n-1}| = 2*8 = 16. For product -33, the minimal |sum| is 14, leading to |a_{n-1}| = 28. So 16 is better.
But wait, maybe there are other factorizations with more roots that give a smaller sum. Let me check.
For example, for product 33, maybe with 5 roots. Let me think. How can I split 33 into 5 distinct integers?
Well, 33 is 1×(-1)×3×(-3)×11. But that product is 1×(-1)×3×(-3)×11 = 99. Not 33. If I want product 33, maybe 1×(-1)×3×(-11)× something. Let me see. 1×(-1)×3×(-11)×x = 33x. To get 33, x must be 1. But duplicate 1. Not allowed.
Alternatively, 1×(-1)×3×(-11)×x = 33x. If x is -1, then product is -33. So that's for product -33. Not helpful.
Alternatively, maybe 1×(-1)×3×(-11)× something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and something else. Wait, but that's the same as before.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, 11, and -1. Duplicate again.
Alternatively, maybe 1, -1, 3, 11, and something else. Let me think. For example, 1, -1, 3, 11, and - something. Let me think. If I have 1, -1, 3, 11, and -x. Then the product is 1×(-1)×3×11×(-x) = 33x. To get 33, x = 1. Then roots would be 1, -1, 3, 11, -1. Duplicate -1. Not allowed.
Alternatively, maybe 1, -1, 3, 11, and something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something. Not helpful.
Alternatively, maybe using more numbers. Let me think. For example, 1, -1, 3, -11, and then 1. Duplicate. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then -1. Duplicate.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
So, seems like for product 33, with 4 roots, the minimal sum is 8. Let me check if there's another way to split into more factors.
Wait, maybe using more factors but with 1, -1, and other numbers. For example, 1, -1, 3, -3, 11. Product is 1×(-1)×3×(-3)×11 = 99. Not 33.
Alternatively, 1, -1, 3, -3, something. Let me think. If I have 1, -1, 3, -3, x. Then product is 1×(-1)×3×(-3)×x = 9x. To get 33, x = 33/9 = 11/3. Not integer. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and something else. But that's 4 numbers. If I add another number, say, 1, -1, 3, -11, and 1. Duplicate again.
Alternatively, maybe 1, -1, 3, -11, and - something. Let me think. Suppose we have 1, -1, 3, -11, -x. Then product is 1×(-1)×3×(-11)×(-x) = -33x. To get 33, -33x = 33 => x = -1. Then roots are 1, -1, 3, -11, -(-1) = 1. So duplicate 1. Not allowed.
So, seems like no way to get more factors with product 33 without duplicates.
Alternatively, maybe using more negative numbers. For example, 1, -1, -3, -11. Product is 1×(-1)×(-3)×(-11) = -33. Which is for the product -33. So, that's not helpful for our case.
Alternatively, maybe 1, -1, -3, 11. Product is 1×(-1)×(-3)×11 = 33. Which is for the product 33. Sum is 1 -1 -3 +11 = 8. So that's the same as before.
So, that's the minimal for product 33. So, sum is 8, absolute value 8, leading to |a_{n-1}| = 2*8 = 16.
But wait, let me think again. Are there other factorizations with different numbers?
For example, maybe using more numbers. Let me think. Suppose we have 1, -1, 3, -11, and then another factor. But that's not working. Alternatively, maybe 1, -1, 3, -11, and something else. Let me think. If I take 1, -1, 3, -11, and then 1. Duplicate. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then -1. Duplicate again.
Alternatively, maybe 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and then something else. If I take 1, -1, 3, -11, and then something else. Let me think of another integer. Let's say 2. Then the product is 1×(-1)×3×(-11)×2 = 66. Not 33. So if I want 33, maybe divide by 2. But that's not integer.
Alternatively, maybe 1, -1, 3, -11, and then something else. If I take 1, -1, 3, -11, and then - something. Let me think. For example, 1, -1, 3, -11, and then - something. Let me think of 1, -1, 3, -11, and then - something. Let me suppose that the product is 33. So, 1×(-1)×3×(-11)×x = 33x. So, 33x = 33 => x=1. But that's duplicate. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then -1. Duplicate again.
So, seems like no other way. Therefore, the minimal sum is 8. So, |a_{n-1}| is 16.
But wait, maybe there are other factorizations with different numbers. Let me think. For example, maybe using more factors. For example, 1, -1, 3, -3, 11. Product is 1×(-1)×3×(-3)×11 = 99. Not 33. But if I take 1, -1, 3, -3, and something else. Let me think. If I have 1, -1, 3, -3, and x. Then product is 1×(-1)×3×(-3)×x = 9x. So, to get 33, x = 33/9 = 11/3. Not integer.
Alternatively, maybe 1, -1, 3, -11, and something. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then 1. Duplicate.
Alternatively, maybe using 1, -1, 3, -11, and then - something. Not helpful.
Alternatively, maybe 1, -1, -3, 11. Which is 4 numbers, product 33, sum 8.
Alternatively, maybe 1, -1, 3, -11. Same thing.
Alternatively, maybe 1, -1, 3, -11, and then another factor. Not helpful.
Alternatively, maybe using more numbers but with 1, -1, 3, -11, and then another number. But that seems not possible.
Alternatively, maybe using 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and something else. Let me think. Suppose we have 1, -1, 3, -11, and then something else. Let me think of 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and then something else. Let me think of 1, -1, 3, -11, and then something else. If I take 1, -1, 3, -11, and then something else. Let me think of 1, -1, 3, -11, and then something else. If I take 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and then something else. If I take 1, -1, 3, -11, and then something else. Let me think of 1, -1, 3, -11, and then something else. If I take 1, -1, 3, -11, and then something else. Maybe something else is something like - something. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Maybe 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and then 1. Duplicate. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then -1. Duplicate.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
So, seems like the minimal sum is 8. Therefore, |a_{n-1}| is 16.
But wait, let me think again. What if the polynomial has more roots? For example, maybe with more than 4 roots. Let me think.
Suppose we have more roots, but with some of them being 1 and -1. For example, let's think of the product as 33. Suppose we have roots 1, -1, 3, -3, 11. Product is 1×(-1)×3×(-3)×11 = 99. Not 33. If we take 1, -1, 3, -3, and something else. Let me think. If I have 1, -1, 3, -3, x. Then product is 9x. So to get 33, x = 11/3. Not integer. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and something else. Let me think. For example, 1, -1, 3, -11, and then 1. Duplicate.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Let me think of another integer. For example, 2. Then product is 1×(-1)×3×(-11)×2 = 66. Not 33. If I want 33, maybe divide by 2. Not integer. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then something else. Let me think. If I take 1, -1, 3, -11, and then - something. Let me say -x. Then product is 1×(-1)×3×(-11)×(-x) = -33x. To get 33, -33x = 33 => x = -1. Then roots are 1, -1, 3, -11, -(-1) = 1. Duplicate. Not allowed.
So, no. Therefore, seems like 8 is the minimal sum for product 33.
But wait, let me think of another possible factorization. For example, 33 can be written as 3×11. So, two factors. Then sum is 3 + 11 = 14. Which is worse than 8. Alternatively, if we have two factors: 33 and 1. Sum is 34. Worse. Or -1 and -33. Sum is -34. Worse.
Alternatively, three factors: 1, 3, 11. Sum 15. Worse than 8. So, 8 is better.
Alternatively, maybe with more factors. For example, 1, -1, 3, -11. Sum 8. Which is better.
Alternatively, maybe 1, -1, -3, 11. Sum is 1 -1 -3 +11 = 8. Same.
Alternatively, 1, -1, 3, -11. Same.
Alternatively, maybe 1, -1, -3, 11. Same.
Alternatively, maybe 1, -1, -11, 3. Same.
So, same sum.
Alternatively, maybe 1, -1, 3, -11. Sum is 1 -1 +3 -11 = -8. Absolute value 8.
So, seems like that's the minimal.
But wait, what if we have more factors. Let me think. For example, 1, -1, 3, -11, and something else. But we saw that it's not possible.
Alternatively, maybe using more numbers. For example, 1, -1, 3, -11, and then something else. But not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and then something else. Let me think of 1, -1, 3, -11, and then something else. Let me think of 1, -1, 3, -11, and then something else. If I take 1, -1, 3, -11, and then something else. Let me think of 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and then something else. If I take 1, -1, 3, -11, and then something else. Let me think of 1, -1, 3, -11, and then something else. Maybe 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and then something else. If I take 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and then something else. Maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Let me think of 1, -1, 3, -11, and then something else. If I take 1, -1, 3, -11, and then something else. For example, 1, -1, 3, -11, and then something else. Maybe 1, -1, 3, -11, and then something else. Not helpful.
So, seems like the minimal sum is 8. Therefore, the minimal |a_{n-1}| is 16.
But wait, let me think if there's another way. What if the polynomial has more roots? For example, maybe with 5 roots. Let me think. For example, if the product is 33, and I have 5 distinct integers. Let me think.
For example, 1, -1, 3, -11, and something else. Let me think. Suppose the product is 1×(-1)×3×(-11)×x = 33x. To get 33, x must be 1. But duplicate. Not allowed. If x is -1, product is -33. So not helpful.
Alternatively, maybe 1, -1, 3, -11, and something else. Not helpful.
Alternatively, maybe 1, -1, 3, -3, something. Product is 1×(-1)×3×(-3)×x = 9x. To get 33, x = 11/3. Not integer.
Alternatively, maybe 1, -1, 3, -11, and something else. Not helpful.
Alternatively, maybe 1, -1, -3, 11, and something else. Product is 1×(-1)×(-3)×11×x = 33x. To get 33, x = 1. Duplicate. Not allowed.
Alternatively, maybe 1, -1, -3, 11, and something else. Not helpful.
Alternatively, maybe 1, -1, -3, 11, and something else. Let me think. If x is -1, duplicate. Not allowed.
Alternatively, maybe 1, -1, -3, 11, and something else. Not helpful.
Alternatively, maybe 1, -1, -3, 11, and something else. Let me think. If I take 1, -1, -3, 11, and then something else. For example, 1, -1, -3, 11, and something else. Let me think of something else. For example, 1, -1, -3, 11, and something else. If I take 1, -1, -3, 11, and then something else. Let me think of 1, -1, -3, 11, and then something else. For example, 1, -1, -3, 11, and then something else. If I take 1, -1, -3, 11, and then something else. Let me think of 1, -1, -3, 11, and then something else. For example, 1, -1, -3, 11, and then something else. Let me think of 1, -1, -3, 11, and then something else. Let me think of 1, -1, -3, 11, and then something else. If I take 1, -1, -3, 11, and then something else. For example, 1, -1, -3, 11, and then something else. If I take 1, -1, -3, 11, and then something else. For example, 1, -1, -3, 11, and then something else. Let me think of something else. For example, 1, -1, -3, 11, and then something else. If I take 1, -1, -3, 11, and then something else. Let me think of something else. For example, 1, -1, -3, 11, and then something else. If I take 1, -1, -3, 11, and then something else. If I take 1, -1, -3, 11, and then something else. Let me think of 1, -1, -3, 11, and then something else. For example, 1, -1, -3, 11, and then something else. If I take 1, -1, -3, 11, and then something else. If I take 1, -1, -3, 11, and then something else. For example, 1, -1, -3, 11, and then something else. If I take 1, -1, -3, 11, and then something else. For example, 1, -1, -3, 11, and then something else. Not helpful.
Alternatively, maybe 1, -1, -3, 11, and then something else. Not helpful.
So, seems like no way to get more roots. Therefore, the minimal sum is 8, leading to |a_{n-1}| = 16.
But wait, let me think again. What if the polynomial has more roots, but with some of them being 1 and -1, and other numbers. For example, let's think of 1, -1, 3, -3, 11. Product is 1×(-1)×3×(-3)×11 = 99. Not 33. But if I take 1, -1, 3, -3, and something else. Let me think. If I have 1, -1, 3, -3, and x. Then product is 9x. To get 33, x = 33/9 = 11/3. Not integer. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and something else. Not helpful.
So, I think I have to conclude that the minimal |a_{n-1}| is 16.
But wait, let me think again. What if the polynomial has more roots, but with some of them being 1 and -1 and other numbers. For example, 1, -1, 3, -11, and then maybe another pair of numbers. But how?
Alternatively, maybe using 1, -1, 3, -11, and then another pair of numbers whose product is 1. But since they have to be distinct integers, the only possible pair is 1 and 1, but duplicates. Not allowed. Or -1 and -1. Duplicates. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
So, seems like no way.
Alternatively, maybe using more numbers with 1 and -1. For example, 1, -1, 3, -11, and then 1. Duplicate. Not allowed.
Alternatively, 1, -1, 3, -11, and then -1. Duplicate.
Alternatively, 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
So, I think the minimal is indeed 16.
But wait, let me check if there is another factorization. For example, maybe with 1, -1, 3, -11, and then something else. Wait, what if the polynomial has more roots but with some of them being 1 and -1 and other numbers. For example, 1, -1, 3, -11, and then 1. Duplicate. Not allowed.
Alternatively, 1, -1, 3, -11, and then -1. Duplicate.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
So, seems like no. Therefore, the minimal is 16.
But wait, let me think of another case. What if the polynomial has more roots? For example, if the product is -33, and we have more roots. For example, 1, -1, 3, -3, 11. Product is 1×(-1)×3×(-3)×11 = 99. Not -33. If I want -33, maybe 1, -1, 3, -3, -11. Product is 1×(-1)×3×(-3)×(-11) = -99. Not helpful.
Alternatively, 1, -1, 3, -11, and something else. Product is 1×(-1)×3×(-11)×x = 33x. To get -33, x = -1. So, roots are 1, -1, 3, -11, -1. Duplicate -1. Not allowed.
Alternatively, 1, -1, 3, -11, and then - something. For example, 1, -1, 3, -11, -x. Product is 1×(-1)×3×(-11)×(-x) = -33x. To get -33, -33x = -33 => x = 1. So roots are 1, -1, 3, -11, -1. Duplicate -1. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and something else. Not helpful.
So, no.
Therefore, the minimal |a_{n-1}| is 16.
But wait, let me check if there are other factorizations for product 33. For example, maybe using 3, -1, 11, -1. But that's same as 1, -1, 3, -11. Sum is 8.
Alternatively, maybe using 3, -1, 11, -1. Same thing.
Alternatively, maybe using 3, -1, -11, 1. Same thing.
Alternatively, maybe using 3, -1, -11, 1. Same thing.
Alternatively, maybe using 11, -1, 3, -1. Same thing.
So, same sum.
Alternatively, maybe using other combinations. For example, 3, 11, -1, -1. But duplicates. Not allowed.
Alternatively, 3, 11, -1, -1. Not allowed.
Alternatively, 3, 11, -1, -1. Duplicate -1.
Alternatively, 3, 11, -1, -1. Not allowed.
Alternatively, 3, 11, -1, -1. Not allowed.
Alternatively, 3, 11, -1, -1. Not allowed.
So, no.
Therefore, I think that the minimal |a_{n-1}| is 16.
But wait, let me think again. What if the polynomial has more roots? For example, if we have 1, -1, 3, -11, and then another root. But we can't. So, maybe the minimal is indeed 16.
Alternatively, maybe there is another set of roots with product 33, but with a smaller sum. For example, let's think of 1, -1, 3, -11. Sum is 8. What if we take 1, -1, 3, -11, and then another root that is 1? But duplicate.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe using 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe using 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
So, I think the answer is 16.
Wait, but let me check another possibility. What if the roots are not just 1, -1, 3, -11, but something else?
For example, suppose the roots are 3, -1, 11, -1. But that's duplicate -1. Not allowed.
Alternatively, 3, -1, 11, -1. Not allowed.
Alternatively, 3, -1, 11, - something. Let me think. For example, 3, -1, 11, - something. If I take 3, -1, 11, - something. The product is 3×(-1)×11×x = -33x. To get 33, -33x = 33 => x = -1. So, roots are 3, -1, 11, -1. Duplicate -1. Not allowed.
Alternatively, 3, -1, 11, - something. Not helpful.
Alternatively, maybe 3, -1, 11, - something. Not helpful.
Alternatively, maybe 3, -1, 11, - something. Not helpful.
Alternatively, maybe 3, -1, 11, - something. Not helpful.
Therefore, no.
Alternatively, maybe 3, -1, -11, 1. Which is same as before.
So, same sum.
Alternatively, maybe 3, -1, -11, 1. Same sum.
Therefore, the minimal is 8, leading to 16.
But wait, let me check if there is a way to have a smaller sum. For example, if the roots are 1, -1, 3, -11, sum is 8. What if we have roots like 1, -1, 3, -11, and then another root that is something else. But no, as discussed before.
Alternatively, maybe roots are 1, -1, 3, -3, and 11. Product is 1×(-1)×3×(-3)×11 = 99. Not 33. Not helpful.
Alternatively, maybe 1, -1, 3, -3, and something else. Product is 9x. To get 33, x is 11/3. Not integer.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Therefore, I think the minimal is indeed 16.
But wait, let me think of another approach. The problem says the polynomial has integer coefficients, distinct integer roots. So, the roots are integers, and the polynomial is 2(x - r_1)...(x - r_n). The constant term is 2 * (-1)^n * product(r_i) = 66. So, product(r_i) = 66 / 2 * (-1)^n = 33 * (-1)^n. So, either product(r_i) = 33 or -33, depending on n.
So, the product of roots is ±33. So, we need to find distinct integers whose product is ±33, and then find the minimal |sum of roots|.
So, the minimal |sum of roots| is 8, as we found, leading to |a_{n-1}| = 2*8 = 16.
But wait, is there a case where the sum is smaller?
For example, if we have more roots, but with more cancellation. For example, let's think of roots that are 1, -1, 3, -3, 11. Product is 1×(-1)×3×(-3)×11 = 99. Not 33. Not helpful.
Alternatively, if we have roots 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe roots 1, -1, 3, -11, and then 1. Duplicate. Not allowed.
Alternatively, maybe roots 1, -1, 3, -11, and then -1. Duplicate.
Alternatively, maybe roots 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe roots 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe roots 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe roots 1, -1, 3, -11, and then something else. Not helpful.
Alternatively, maybe roots 1, -1, 3, -11, and then something else. Not helpful.
Therefore, I think the minimal is indeed 16.
But wait, let me think of another case. For example, what if the polynomial has roots that are 1, -1, 3, -11, and then another pair of numbers that multiply to 1. But since they must be distinct integers, the only way to have product 1 is 1 and 1 or -1 and -1. But duplicates. Not allowed.
Alternatively, maybe 1, -1, 3, -11, and then something else. Not helpful.
Therefore, I think the answer is 16.
But wait, I just thought of something. What if the polynomial has more roots, but with some of them being 1, -1, 3, -3, and then something else? Let me think. For example, if the product is 33. Suppose we have roots 1, -1, 3, -3, and x. Product is 1×(-1)×3×(-3)×x = 9x. To get 33, x = 11/3. Not integer. Not allowed.
Alternatively, if the product is -33, then 9x = -33 => x = -11/3. Not integer.
Alternatively, maybe roots 1, -1, 3, -3, 11. Product is 99. Not helpful.
Alternatively, maybe roots 1, -1, 3, -3, -11. Product is -99. Not helpful.
Alternatively, maybe roots 1, -1, 3, -3, something else. Not helpful.
Therefore, no.
So, I think the minimal |a_{n-1}| is 16.
But let me check if there are other possible factorizations. For example, maybe using 33 as a single root. Then, the polynomial would be 2(x - 33). But then the constant term is 2*(-33) = -66. But the problem says the constant term is 66. So, that's not allowed. Therefore, the product of the roots must be 33 or -33, and with the leading coefficient 2.
Alternatively, if the polynomial has two roots. For example, 2(x - a)(x - b). Then the constant term is 2ab = 66 => ab = 33. So, a and b are integers, distinct, and product 33. So possible pairs are (1,33), (3,11), (-1,-33), (-3,-11), ( -1, -33), etc.
Then the sum of roots is a + b. So, for (1,33): sum is 34. |a_{n-1}| = 2*34 = 68.
For (3,11): sum is 14. |a_{n-1}| = 28.
For (-1, -33): sum is -34. |a_{n-1}| = 68.
For (-3, -11): sum is -14. |a_{n-1}| = 28.
So, minimal is 28 here.
But earlier, with four roots, we had |a_{n-1}| = 16, which is better.
So, better.
So, with two roots, minimal is 28.
With three roots, for example, product 33. Let's see. Three distinct integers with product 33. Possible options: 1, 3, 11. Sum is 15. |a_{n-1}| = 2*15 = 30. Worse than 16.
Or for product -33, three distinct integers. For example, 1, -1, -33. Sum is 1 -1 -33 = -33. |a_{n-1}| = 66. Worse.
Or 1, -3, 11. Product is 1×(-3)×11 = -33. Sum is 1 -3 + 11 = 9. |a_{n-1}| = 18. Worse than 16.
So, worse.
With four roots, we have sum 8, leading to 16.
So, better.
With five roots, as discussed before, no.
So, the minimal is 16.
Therefore, the answer should be 16.
But wait, let me check if there are other four-root combinations. For example, 1, -1, 3, -11. Sum 8. What about other four roots?
For example, 1, -1, -3, 11. Sum is 1 -1 -3 + 11 = 8. Same.
Or 1, -1, 3, -11. Same.
Or 1, -1, -3, 11. Same.
Or 3, -1, 11, -1. Duplicate -1. Not allowed.
Or 3, -1, 11, - something. Not helpful.
So, no other four-root combinations with product 33.
Alternatively, maybe other four-root combinations with product 33. For example, 1, -1, 3, -11. Which is same as above.
Alternatively, maybe 1, -1, 11, -3. Same.
Alternatively, maybe 1, -1, 3, -11. Same.
So, same sum.
Therefore, the minimal |a_{n-1}| is 16.
But wait, let me check one more thing. What if the polynomial has more roots, but with some of them being 1, -1, 3, -11, and then another root that is something. For example, maybe 1, -1, 3, -11, and something else. But as discussed, it's not possible.
Alternatively, maybe 1, -1, 3, -11, and something else. Not helpful.
Therefore, I think the answer is 16.
**Final Answer**
\boxed{16}
</think>
To solve the problem, we are given a polynomial:
$$
f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0
$$
with **integer coefficients**, **distinct integer roots**, and the conditions:
- $ a_n = 2 $
- $ a_0 = 66 $
Our goal is to **minimize** the value of $ |a_{n-1}| $.
---
### Step 1: Use the Structure of the Polynomial
Given that the polynomial has **integer coefficients** and **distinct integer roots**, we can write it as:
$$
f(x) = 2(x - r_1)(x - r_2)\cdots(x - r_n)
$$
This is because the leading coefficient is $ a_n = 2 $, and the roots are integers.
---
### Step 2: Use Vieta's Formulas
From Vieta's formulas:
- The **constant term** is:
$$
a_0 = 2 \cdot (-1)^n \cdot r_1r_2\cdots r_n = 66
\Rightarrow (-1)^n \cdot r_1r_2\cdots r_n = 33
$$
So the product of the roots $ r_1r_2\cdots r_n = \pm33 $, depending on whether $ n $ is even or odd.
- The **coefficient of $ x^{n-1} $** is:
$$
a_{n-1} = -2 \cdot (r_1 + r_2 + \cdots + r_n)
\Rightarrow |a_{n-1}| = 2 \cdot |r_1 + r_2 + \cdots + r_n|
$$
So, our task is to find a set of **distinct integers** whose product is $ \pm33 $, and whose **sum** has the **smallest absolute value**.
---
### Step 3: Try Integer Factorizations of $ \pm33 $
We want distinct integers whose product is $ \pm33 $, and the **sum of those integers** is as small as possible in absolute value.
Let’s consider:
- $ 33 = 1 \cdot 3 \cdot 11 $
- $ -33 = (-1) \cdot 3 \cdot 11 $
But these are only three distinct integers. Let’s try to find a set with more elements (i.e., more roots) that still multiply to $ \pm33 $, but whose **sum is small**.
Try the set:
$$
\{1, -1, 3, -11\}
$$
- Product: $ 1 \cdot (-1) \cdot 3 \cdot (-11) = 33 $
- Sum: $ 1 - 1 + 3 - 11 = -8 $
Thus, $ |a_{n-1}| = 2 \cdot | -8 | = 16 $
---
### Step 4: Check for Smaller Sums
Try other combinations of integers that multiply to $ \pm33 $, but with more distinct elements. For example:
- $ \{1, -1, -3, 11\} $: product = $ 33 $, sum = $ 1 - 1 - 3 + 11 = 8 $
- $ \{1, -1, 3, -11\} $: same as above
No other set of distinct integers with more than four elements can multiply to $ \pm33 $ without including duplicate values or non-integer values.
Thus, the **smallest possible** absolute value of the sum of the roots is **8**, and the corresponding value of $ |a_{n-1}| $ is:
$$
|a_{n-1}| = 2 \cdot 8 = 16
$$
---
### Final Answer
$$
\boxed{16}
$$